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What do the number at end of model numbers mean?
When i see thing on the internet about aircrafts it often has a number at the end of the model number such as B737-300, what does the number after the dash mean?

Absolutely NOT a duplicate.  This question asks what the -xxx MEANS, the other question asks how the numbers of the -xxx are CHOSEN.

In the case of the 737, the first series produced was the 737-100.  Not many were made.
Then the 737-200 was built, which was very popular.
The third generation of the 737, known now as the "Classic", had three variants, the -300, the -400, and the -500.  All are very similar, the significant difference being the size.  The -300 has bigger (more powerful, as well as physically larger) engines than the older -200 did, and a longer cabin.  With its success, Boeing decided to lengthen the cabin further, resulting in the -400.  They also made a variant that was essentially the length of the shorter -200, and since it came along third it got the next designation, the -500.
The next generation of 737's, known as the "Next Generation", runs from the -600 through the -900, and this time Boeing decided to make the numbers correspond with the fuselage length, so the -600 is about as long as the -200 and -500 (roughly), the -700 is comparable to the -300, the -800 is longer as the -400 was, and the -900 is longer than anything before.
Now they're out of numbers, and so the NEXT generation of 737's, known as the Max, gets single digit numbers... the Max -7 (i.e. 737-7) corresponds roughly to the length of the -700, the Max -8 to the -800, the Max -9 to the -900, and so on.
The point of this long discussion is simply that the numbers following the dash represent different variants of the aircraft.  In some cases, it may be a new generation design; in other cases it may simply be a different fuselage length.
As the suggested-as-duplicate question discusses, HOW those numbers are chosen is sometimes driven more by marketing than it is by mathematical progression.  Often, though, the higher number indicates a later version, and often-but-not-always, a newer generation.  But you really have to know the particular aircraft line to know which it is.

It might be interesting to mention that there is one type, A320, where the variants modify the main number. A319 is to A320 as 737-500 is to 737-300 (but then they still have the variant number and A321-200 is to A321-100 about as 737-9 is to 737-900, i.e. same size, upgraded design).

Also, Airbus did not repeat this with the other models. A330 has -200 and -300 be the different sizes.

It should be noted that -700 (or -600, -800, etc) is a model series with a basic configuration.  Each example of the aircraft will have a different model number if it differs from other variants. So there are a large number of different model numbers.  From the Type Certificate Data Sheet (TCDS) A16WE, the -700 series starts with the -705, -724, -732, -752...   TCDS are available on the FAA website here.

The number after the dash is basically just the variation of that aircraft type. For example we have the Boeing 777-300, however there is a newer model coming out called the Boeing 777-900.
It just helps differentiate the different models of that aircraft types.

welcome to av.se!

Is it really -900? I thought it was just 777-9?


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What are the differences between wings made for 2 engines and wings for 4 engines?
I have always wondered the differences between a wing made to hold 2 engines and a wing made to hold 4 engines, could someone please explain?

Hum... what do you want to know? It is about why the engine number varies from one aircraft to the other... and possibly whether wings need to be adjusted according to the number of engines; or really about possible basic difference in the wing, regardless of the reason of 2 or 4 (or even more) engines?

I don't have an answer for this, but some relevant information may be available around the wings for A330 and A340. They differ mostly be the number of engines, and the differences in the wings will be large due to this difference.

@Monolo -- actually, the A330 and A340 use the same wing structure -- in the A330's case, its simply missing the outboard pylons and engines.  (This came in handy when the A330 MRTT was developed as the outboard hardpoints were reused for refueling pods)

@mins Or fewer -- most wings hold zero, one or two engines. ;-) :-P

The differences are surprisingly small, and the implications of two versus four engines are not very great as far as design of the wing is concerned.
Weight implications
The wing is an immensely strong structure - after all, it's able to bear the weight of the entire aircraft, including of course the engines.
Engines are heavy, but only a small fraction of the weight of the whole aircraft.
A similar aircraft with four engines rather than two will have smaller engines, so the total weight difference will not be large.
Space implications
You need more space to fit two engines on each side, but not a vast amount. For example, the British Aerospace 146 has four remarkably small Lycoming ALF 502 engines. What's more, four small engines can be easier to mount, because they need less ground clearance.
Same wings, more engines
In practice, it's not very difficult to add engines to a wing (for aircraft companies, that is; I don't think I'd be able to do it without seeing someone do it first at least once).
For example, the 747 can optionally carry a fifth engine - obviously, this engine is just being carried, not used, but it shows that simply placing the engine there doesn't present insurmountable aerodynamic or other engineering difficulties.
And then we have the Airbus A330/A340, designed and built in parallel, which are essentially the same aircraft - same fuselage and same wing - one with two engines and one with four. (Actually the -500 and -600 A340 variants do have slightly longer wings, but that's because they are heavier, and nothing to do with the additional engines.)
That's not to say that there are no implications in having additional engines. In this case, wing flutter caused by the outboard engines of the A340 required some stiffening.
And of course, additional engines will require additional fuel and control lines. However, these aren't part of the structure of the wing itself, and are not in themselves likely to have implications for it.

I think the weight implications might be larger to you suggest, engines on the wings can cause quite some bending moment relief (thereby lightening the wing structure)

The A340-500/600 also has a different sweep angle and also a twist that the A340-200/300 and A330 doesn't have, but otherwise you are correct.  The only real implication an additional engine has on the wing is actually the ability to carry more fuel in the outer wing tanks, because the bending moment is damped by the extra engine...

Wings on commercial airplanes are complex, they have electrical lines, hydraulics, fuel tanks, lines and pumps plus structure to handle the forces involved in flight. A wing that has engines on it will need additional structure for the engine to mount on, and have to be stronger in some ways because of the weight of the engine and the thrust it produces.  So a wing that has engines on it will need to be stronger than one without, and therefore will be heavier and possibly thicker at points.
The difference between 2 engines and 4 is probably less than having no engines and 2 engines, but there are considerations. The weight of the outer engine will have a bigger moment arm than the inner engine, so the wing will need to be stronger to compensate. The additional engine controls and fuel cross-feed lines will need more space in the wing than a wing with 1 engine, perhaps making the wing thicker or bigger than it would be on one with 1 engine.

In many respects a wing with an engine on it is stressed less than if the same engine was mounted elsewhere. In flight, an engine on the wing is right where lift is generated, so the force to carry it does not need to be transmitted all the way to the fuselage (as would be the case for a body-mounted engine). On the ground, a wing-mounted engine must be carried by the wing, of course, but this is still less load than the fact that when airborne the wing carries the entire fuselage! (Though in the latter case the bending moment goes in the opposite direction).

@HenningMakholm you're correct that the engine mass is quite helpful in the wing root bending moment. The joints and surrounding structure is much affected by the thrust and crash landing forces of the engines. But still a distributed load on the wing is always a good thing, which must be a trade off against the added complexity, number of serviced elements, cost, economies etc.

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Would a perfectly silent supersonic aircraft create a sonic boom?
Assume that you have an aircraft that is so amazingly stealthy that it is perfectly quiet: no engine noise at all, nothing.  And, it's going supersonic.  (Maybe in a steep dive with the motor shut off.)  Will it make a sonic boom?
Would the sonic boom be louder if the aircraft was making more noise but all else remained constant?
The (excellent) answer to this question discusses aircraft size in relation to the strength of the sonic boom, but no mention of aircraft noise.  So would the amount of noise the aircraft makes matter?
Edit: by "silent" I'm not trying to imply "displaces no air" or creates no sound by its passage.  The point is simply that the aircraft itself doesn't make any noise from an engine or otherwise.  It's silent, like a bullet, or a glider.

A totally silent aircraft is probably so unphysical that this question has no real basis for an answer (it's like asking about transonic control authority on a magic carpet)

Does a supersonic bullet create a sonic boom?  (Hint: yes, it does, although it's more of a click you hear before the report of the gun being fired.) (I don't recommend getting into a position where you can observe this first-hand, if you can avoid it!) So: does the sonic boom get louder as the bullet/aircraft/whatever gets louder?

Depends on what you mean by "silent" and "aircraft".  The "Busemann's Biplane" airfoil configuration, for example, doesn't produce a sonic boom, but it doesn't produce lift, either.

But subsonic bullets aren't silent. If it is perfectly silent, then by definition there is no pressure wave created. That is a totally unphysical assumption, and makes analogies to any real object highly questionable.

You need to send this question over to XKCD, in his "What If" section.  Very very interesting hypothetical question.  Not sure how you'd make a soundless craft but...  Still.

Sonic booms aren't created by the aircraft's engine noise. They're created by the aircraft moving through the air. The aircraft has a non-zero volume (an aircraft with zero volume is like the emperor's new clothes; you can argue all you want how much design and engineering work it took but all anyone will see is an empty spot on the flight line), meaning it must displace air as the plane passes through air. This displaced air equalizes with the surrounding air in an expanding pressure wave, which moves at the speed of sound in all directions.
As your hypothetical silent aircraft approaches the speed of sound, the air being pushed aside can no longer equalize by moving out directly in front of the aircraft. This causes a single strong compression wave to build in front of the aircraft. Similar waves have already started to build over the top of the wing surfaces, as the airflow over the leading edge of the wing (by design intended to make the air move faster thus lowering its pressure and producing lift) is temporarily supersonic before friction causes it to settle back below sonic speed; at this transition point between super- and sub-sonic, air still moving supersonically "backs up" against the subsonic airflow, creating a wall of relatively high-density air.
At this point, an observer would already hear a small sonic boom from some combination of these compression waves. However, the boom becomes much more pronounced as the entire aircraft transitions beyond Mach 1. At this point the "wall" in front of the aircraft becomes more of a "cone" as the air can no longer build up directly in front of the jet and is instead violently pushed aside at any local "disconnect" in the angle presented to the oncoming airflow. This sudden compression produces a strong shockwave "wake" which is what is heard as a sonic boom.
As a perfect example of this, there actually is (was) a vehicle that spends a very critical portion of its flight under zero engine power, effectively silent in that regard, at not only supersonic but hypersonic speeds. This vehicle, as shown in the linked YouTube, produces a characteristic "double sonic boom", proof that even a "silent" aircraft isn't silent when it's supersonic. Can you guess what it is?

The noise of an aircraft's engines could, conceivably, add a little extra to the sonic boom; the biggest difference is that immediately following the sonic boom, if you were close enough, you'd also get the engine noise, which can be deafening all by itself at full throttle up to 100 feet away.

Keep in mind, the question is extremely hypothetical...  Pushing air aside would create pressure waves, which are (by definition) what sound waves are.  So a noiseless craft would have to (somehow, and no I don't know how) somehow it would have to not displace the air.  Otherwise sound would be produced.  So, while I agree that your scenario is likely (if we follow those silly laws of physics), if you somehow created a noiseless aircraft (one that doesn't displace air), I'm not sure this would apply. :)

By "silent" I was thinking of a hypothetical machine that could fly under power without engine noise or any related thrust turbulence noises. Gliders would be the only such craft. I was going to say no glider could be supersonic, but there actually is a very famous vehicle that glides with zero thrust power at supersonic speeds that is a perfect example.

So, would the sonic boom be louder (or otherwise different) if the aircraft instead of having no engine noise had a whole lot of engine noise?

Ah, and the question was clarified as well to say that your reading of it was the correct one.  So, yes, you'd be right.

Gliders are far from silent.

@Simon In order to answer this extremely hypothetical question, one has to come up with a definition of "silent". Keith has chosen to define "silent" as meaning that the only sound generated by the aircraft is from its movement through the air, i.e., that the propulsion mechanism makes no sound. A glider is clearly silent in that sense.

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Long pole on the tail of a Cessna, what could it be?
I live just near the finals for our local airport. Today while in the backyard I saw a Cessna fly overhead for landing, it was a tricycle gear and looked like a 150/152, but may have been a 172.
There was a long pole that appeared to be coming out of the tail, although there is a small chance it could have been from the wing on the other side of the aircraft from me. It wasn't very wide in diameter (I couldn't tell even an estimate), was maybe around two-thirds of the length of the aircraft, was painted the same white as the aircraft and capped at the end by a black piece that appeared to be hemispherical.
Does anybody know what this pole could be?

You didn't happen to take a picture, did you?

Could it be a runway calibrator aircraft?

Are you certain it wasn't a 205/206/207 or a Caravan?

No sorry, I did not take a picture as he did not make another circuit. It looked like a 150 or 172 to me, but could have been a 200. I am certain it was not a Caravan, much smaller.

Backlink to a related newer question, but with photo  https://aviation.stackexchange.com/questions/82481/what-is-this-part-of-an-aircraft-looks-like-a-long-thick-pole-sticking-out-of-t

What you describe does not match any banner tow rig I am familiar with, and I have towed banners.
What it sounds like is a magnetometer, which might be on a survey aircraft. They can be on a boom (stinger) or sometimes towed.  I have seen them on Cessna 182, 210 and 208 aircraft, as tail mounted booms.
Magnetometers are typically used for geophysical survey work.  Towed magnetometers may be more sensitive and might be useful for detecting submarines, mines, etc. The P-3 typically has a rear boom with a magnetometer or magnetic anomaly detector (MAD).

image of one: https://www.lakecentral.com/wp-content/uploads/2016/03/8-IMG_0483-TDH.jpg (obviously not mine, taken from https://www.lakecentral.com/product_sec/c185-survey-tail-boom/)

@Baldrickk, yup.  Most are skinnier tubes, but obviously not all.  Some also articulate, to provide additional tail clearance.  The geophysical guys do allot of grass and dirt strips, and sometimes high DENALT conditions, so the takeoff attitude may be a bit nose high.  Then the boom is lowered in flight, usually with an electric winch inside the plane.

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Does an instrument exist to measure the different types of drag?
Does a 'drag meter' exist that separately shows total, induced and parasite drag?
If it doesn't, and in theory, would it be possible to build one?

Yes, partially.  A wind tunnel's wake rake (an array of pitot tubes) can be adapted to measure in-flight profile drag (form drag plus skin friction).
Measuring Wing Profile Drag using an Integrating Wake Rake, by Pifer and Bramesfeld, in Technical Soaring 36(3):74-82, 2012.
The article doesn't have any in-flight photos, so it may not have actually been flown at time of publication.

@quietflyer mentioned Richard Johnson, who indeed has flown an airplane with a drag rake.

The wing drag rake was installed on N3176T’s left wing trailing edge, 1.0 meters outboard from the wing root joint. ...

R. Johnson, "A flight test evaluation of the SZD-55-1," Soaring, March 1992.
The drag rake itself is described and shown in another article:
R. Johnson, "At last: an instrument that reads drag!," Soaring, October 1983.

JZYL's answer explains how a wake rake works, and suggests that it would help to measure not just pressure but also cross-flow.  But it would be hard to tow an array of cross-flow sensors (hot film, maybe?) without affecting the airplane's drag and center of mass.

@xxavier and others-- Drag rakes have certainly been used in flight to measure the drag of sailplanes-- for examples, see some of Richard Johnson's work.  I am fairly sure the instrument is rigidly mounted to the aircraft, not towed.  This answer could be improved by incorporating references to this work.  In fact I'm fairly sure I saw a glider with a drag rake on it at the soaring museum in Moriarty, New Mexico.  If I can locate photos I may compose an actual answer along these lines.

Drag rakes have been flown behind wings for at least half a century already. However, photographic evidence is missing from the web.

1, no, a rake can't know what caused the change in pressure.  2, great photo, maybe it is a drag rake.  What context is there for this photo?  (This is worth asking as a new question, instead of hidden in these comments.)

In theory, it's possible to distinguish induced drag from parasite drag, at least in subsonic flow. The drag force can be quantified through conservation of momentum in a control volume (note that the coordinate system is in the stability axis):

$$D=-\oint_{S_\infty}p\hat{i}\cdot d\vec{S} - \oint_{S_\infty}\rho (\hat{i}\cdot\vec{u})(\vec{u}\cdot d\vec{S})$$
where $p$ is static pressure and $\rho$ is density.
If the sides of the control volume are far away from the wing/airplane, then the flow field tends to the free-stream values and their contributions go to zero. The only side that matters is the aft boundary, called the Trefftz plane.
After some algebra (Ref. Drela, Flight Vehicle Aerodynamics), we can separate the induced drag and parasite drag (subscript $_\infty$ denotes free-stream values):

Induced drag ($v$ and $w$ are cross-flow components): $$D_i=\int_{S_T}{\rho_\infty \frac{v^2+w^2}{2}dS}$$
Parasite drag ($u$ is axial flow speed, $V_\infty$ is the free-stream speed): $$D_p=\int_{S_T}{\rho u(V_\infty-u)dS}$$

To summarize, you can measure induced and parasite drag contributions by casting a large net aft and far of the wing/airplane (Trefftz plane). To measure induced drag, you would need an array of probes that can measure the cross-flow at individual locations on the Trefftz plane. To measure parasite drag, you would need an array of pitot tubes to measure the total pressure and the resulting momentum defect.

@Сократ v is lateral component, w is vertical

v lateral component is zero if wing is flat?

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Why is the ratio maximum range speed/maximum endurance speed the same for any airplane?
It seems counter-intuitive that the ratio of speeds for maximum range and maximum endurance is always the same, $3^{0.25} = 1.316...$ for any heavier-than-air aircraft, regardless of its mass, size or form. It's hard to believe that such ratio may be identical for an ultralight autogyro and for the F-35 fighter, but so it seems to be...
Why is it so?

Yes... You can find a derivation in https://cafe.foundation/v2/pdf_tech/MPG.engines/AIAA.1980.1847.B.H.Carson.pdf

And it's also here: http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node99.html

Also in page 877 of 'General Aviation Aircraft Design' by Snorri Gudmundsson  [img]http://i.imgur.com/UfkXwOT.png?1[/img]

Comments are second-rate citizens on the SE network. Please add your links into your question, preferably in a nice summation :-)

The ratio is the same for all aeroplanes if you accept a number of assumptions:

The propulsion efficiency is constant, regardless of speed or power setting
Aerodynamic drag is the sum of parasite drag and induced drag
Parasite drag is proportional to the square of airspeed:$ D_p = k_p \cdot V^2$
Induced drag is inversely proportional to the square of airspeed: $D_i = \frac{k_i }{V^2}$
There is no wind

Since we assume the propulsion efficiency is constant, the fuel consumption rate is directly proportional to the power required. The power required is drag times airspeed:
$P = D\cdot V = D_p\cdot V + D_i\cdot V = k_p\cdot V^3 + \frac{k_i}{V}$
For the maximum endurance we need to minimise the fuel consumption and thus we need to find the speed that minimises the power.
$\frac{dP}{dV} = 3 k_p V^2 - \frac{k_i}{V^2} = 0$
Solving for $V$ results in
$V_{endurance} =  \sqrt[\uproot{1}4]{   \frac{k_i}{3k_p}} $
For the maximum range we need to find the speed that minimises the fuel consumption per distance travelled, which is found when the ratio of power to  speed over ground is minimal. As we assume there is no wind, the ground speed and the airspeed are equal.  Since the ratio of power to airspeed is drag, we have to find the speed for minimum drag:
$\frac{dD}{dV} = 2 k_p V - 2\frac{k_i}{V^3} = 0$
Solving for $V$ results in
$V_{range} =   \sqrt[\uproot{1}4]{   \frac{k_i}{k_p}} $
We can now show that the ratio of maximum range speed to maximum endurance speed is:
$\frac{V_{range}}{V_{endurance}} = \left. \sqrt[\uproot{1}4]{   \frac{k_i}{k_p}} \middle/  \sqrt[\uproot{1}4]{\frac{k_i}{3k_p}} \right. = \left. 1 \middle/  \sqrt[\uproot{1}4]{\frac{1}{3}} \right. =  \sqrt[\uproot{1}4]{ 3} = 1.316... $

Now as long as the assumptions hold, for any aircraft the ratio of the speed will be approximately 1.3.
For the F-35 the ratio may be approximately right in the subsonic domain. Effects of compressibility will cause the drag to be higher when the aircraft gets into the transonic domain so depending on the max-range speed the last two assumptions may not hold.
For an autogyro at higher speeds, the relation of induced drag to airspeed is inversely proportional to the square of velocity, just like conventional fixed wing aircraft. This is because the fundamental way of creating lift by deflecting the incoming airflow downwards is the same for autogyros and aeroplanes. Therefore autogyros have also a factor pf approximately 1.3 between their maximum endurance and maximum range speeds.

I thought I was reading a Peter Kampf answer, and believe me, that's a compliment!

Those ratios are only correct if Mach and engine efficiency aren't included in the calculations.  If one considers that best range speed, from your calc, gets you into an unfavorable Mach number then it won't be correct.  If best range speed forces engine operation at a inefficient power setting, (low power settings will be inefficient for turbines, high power settings are inefficient for gasoline engines) then that calculation will be incorrect.
I remember best rate of climb speeds in the T-38 were crazy high, like 540CAS until Mach 0.8 something, even though this was nowhere close to l/d max.  The engines were straight jets, so performance was very good at speeds higher than what was ideal from an airframe perspective.  This is largely why jets must fly so high to get reasonable efficiency, the engines don't lose much thrust with speed, but the airframe needs to be near l/d max CAS to be efficient.

Why is the speed for maximum endurance always lower than the speed for maximum specific range though?

@MBC870 Logically, how could best range speed be slower than best endurance speed? How could it be possible to have better range if traveling both slower and for less time?


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Is this equation for propeller thrust correct?
From dimensional considerations, and assuming that the power $P$ applied to a propeller of diameter $D$ moving at an axial speed $V$ is 100% converted to the acceleration of a mass of air with density $rho$, I have arrived to the following expression for the thrust $F$:
$F=k(P·rho·V)^{1/2}·D$
where $k$ is a non-dimensional constant.
Is that expression basically correct? I'm now editing to include below the derivation, a bit long...
We have a propeller with a diameter $D$, absorbing a power $P$ when moving at an axial velocity $V$ in air of density $\rho$. Let’s assume that there exists a function f such that the thrust $F$ of the propeller is:
$F = f(P,D,V,\rho)$
The variables are thrust $F$, dimensions $MLT^{–2}$; Power $P$, dimensions    $ML^2T^{–3}$; prop diameter $D$, dimensions $L$ and air density $\rho$, dimensions $ML^{–3}$
Five variables are too much. The system can’t be solved unless we have four, one dependent and  tree independent…  So, in place of $V$ and $D$, we take the volume $W$ swept by the spinning propeller in the unit of time, dimensions $L^{3}T^{–1}$
$W = π/4 · D^{2} · V$
There is an non-dimensional constant $k$ such that:
$k = F^a\cdot P^b\cdot \rho^c \cdot W^d$ where $a,b,c,d$ are numbers to be determined.
Switching to the dimensions:
$M^0 L^0 T^0 = (MLT^{–2})^a  (ML^2T^{–3})^b  (ML^{–3})^c (L^3 T^{–1})^d$
Then, the system is:
$0 = a + b + c\\
0 = a + 2b –3c + 3d \\
0 = –2a –3b –d$
$F$ is the dependent variable, so we make a=1
Solving the system:
$b = –1/2 \\
d = –1/2 \\
c = –1/2$
Inserting in
$k = F^a\cdot P^b\cdot \rho^c \cdot W^d$ where $a,b,c,d$
the values of the exponents, and solving for $F$,
$F = k\cdot P^{1/2}\cdot \rho^{1/2} \cdot W^{1/2}$
But $W = π/4 · D^{2} · V$
Now $k$ may absorb the constant π/4, and we get:
$F = k\cdot P^{1/2}\cdot \rho^{1/2} \cdot V^{1/2}\cdot D$

this seems related, and Peter's answer seem to contain what you need: https://aviation.stackexchange.com/questions/8819/is-there-any-equation-to-bind-velocity-thrust-and-power?rq=1

How did you calculate the exponents? I can identify four unknown exponents, but only three dimesional equations.

@Federico In the –quite likely– case that Kämpf's answer is right, the expression I've derived is probably wrong...

@Gypaets Give me some time, and I will write down the derivation...

You already wrote the derivation for the static case (as in Peter's answer) in this question https://aviation.stackexchange.com/a/44654/8749. The point in this question is if there is a thrust dependency on the speed $V$ or not.

@Gypaets Yes, but in this case there's an axial speed $V$, hence it's different...

@Gypaets  Derivation included...

@xxavier.  Thrust is caused by a change in momentum, or/and an increase in static pressure, acting over an area. So, for a gas turbine, Net thrust (FN) = M x  (Ve - Va) / g + Ae x (Pse - Pamb). Va is the aircraft velocity, Ve is the exhaust velocity. The physics is the same for a prop, but when the flow is subsonic, Pse = Pamb. So, you are left with the first term, and your velocity term should be a change in velocity, of the mass flow going through the prop. This confers with the general form of Peter's answer.

@Penguin  A dimensional derivation is independent of the underlying physical facts. The variables have to be chosen well, but that's all... The velocity term that I consider is not that of the moving air mass, but the axial velocity of the prop, with respect to the distant mass of air at rest.

@xxavier. Fully understand the dimensional analysis approach. But you can use DA and an understanding of the physics to arrive an equation which is dimensional correct and physically correct. The two are not mutually exclusive. And... your question is asking if your equation is correct, so I am answering that by showing one of the errors.

@Penguin Yes, you are right pointing out that the physical, underlying reality should not be ignored, but what I want is a derivation of thrust as a function of power absorbed, prop diameter and prop axial velocity with respect to distant, calm air; and not to the velocity imparted by the prop to the air mass...

@xxavier. OK, if that's a constraint you want to include, fair enough. A pity no one has answered yet, I think it's actually one of the more technically interesting and well thought out questions for quite a while.

@xxavier What is prop axial velocity exactly ? Do you mean  flight velocity V0 or air axial velocity in stream tube (Momentum + actuator disk theory) at prop position ?

@Acsed I mean flight velocity...

@xxavier: then, I do not really like the use of your variable « W » which is not the volumetric flow through the propeller.
Moreover, even if your derivation seems correct, why your dimension analysing did not include rotational speed (N) or torque (T) ? Commonly, Thrust and Power are related by overall prop efficiency with « eta = F V0 / Pw » where Power is given by both N and T. Maybe you should try another derivation with one more variable leading to 2 or 3 dimensionless numbers.

@Acsed Thanks for the suggestion. I'll try doing as you say...

From the comments:what I want is a derivation of thrust as a function of power absorbed, prop diameter and prop axial velocity..

Power absorbed $P$ is engine torque  times prop angular velocity $\omega$: $P =  \cdot \omega \tag{1}$.
Prop axial velocity at the tip $V_t =\omega \cdot D \tag{2}$

$P$ is the power applied in the axial direction, while thrust is generated in the direction perpendicular to the propeller plane - there must be some function describing the relationship between the forces in both directions. This function would be one describing efficiency and losses, since the primary force application is the engine torque and the thrust is the net resulting force.
If the goal is like is stated in OP ...propeller of diameter  moving at an axial speed  is 100% converted to the acceleration of a mass of air... this means that there are no losses, and we can use the simple impulse theory of the magic disk. This is comprehensively explained in this post, which results in the following equation between thrust and power:
$$P = \sqrt{\frac{F^3}{2\rho A}} \tag{3}$$
If we want $V_t$ in there we need to substitute (1) and (2):
$$\frac { \cdot V_t}{D} = \sqrt{\frac{F^3}{2\rho A}} \tag{4}$$
Since A = propeller disk area = $\frac {\pi}{4} \cdot D^2$, (4) further boils down to:
$$F^{3/2} = k \cdot  \cdot V_t \cdot \rho^{1/2} \tag{5}$$
Note that (3) contains $P$ but no $V_t$, while (5) contains $V_t$ but no P. We could lump the motor torque into the force $F$ dimension, but all my engineering genes scream to not do that because of the origin and effect relationship: no engine torque, no net thrust.
Both (3) and (5) contain $\rho$, which is a must considering that the thrust is obtained by acceleration of a mass of air. The mass flow dimension can disappear into a force dimension as well when multiplied by a velocity, for instance a propeller tip velocity...
This answer shows the relationship between thrust and power of a helicopter rotor in the hover (and therefore of a propeller) using modified momentum theory with losses:
$$C_P = \frac{\kappa \cdot {C_T}^{3/2}}{\sqrt{2}} + \frac{\sigma \cdot {C_D}_0}{8}$$
with $\kappa$ = loss factor and $\sigma$ = blade solidity. So the exponent of the function (induced power) vs. (thrust) is 3/2. Also, there is an additional term that the dimensional analysis cannot catch, which is power required in the propeller plane.

I see a problem: Equation (3) is derived from momentum considerations for the case of static thrust. that's an special case, since, in that condition, the angles of attack 'seen' by the blades are much greater than the angle 'seen' by the blades when there is a significant axial velocity of the prop disk...


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Which airplane has had the longest production run?
An earlier question received interesting answers on what airplanes have been in service the longest; this question asks about airplanes in production the longest.  By this, I mean from earliest production of the first variant, to the end of the production run of the latest variant.  Thus, the 737 would count from 1967 through the present, even though there have been plenty of variants along the way.
Regarding breaks in production... EDIT: any break in production less than a decade, doesn't rule out an aircraft.  Multiple breaks in production that total less than a decade of "down" time don't either. An aircraft that is essentially the same produced by somebody else is fine if both are generally regarded as being the same aircraft, even if the designation is different (i.e. the F4U was built by multiple manufacturers, but today everybody thinks of any of them as a Corsair).
Military, civilian, large & small, even including gliders & helos (if any have been in production long enough to be competitive) are all fair game.  We can mention balloons (as "aircraft," though not really "airplanes") as having been produced since before the Wright brothers, but that's not really what I have in mind here.  Edit: Nor is "oldest aircraft still in production," even though the answers tend to be similar since it looks like the aircraft mentioned there are still in production.  This question is about length of production run, even if it is complete.
So, longest production run... what do we have?

Related: Oldest aircraft still in production?

Depending on maker some variant of the Piper Cub has been in production since 1938 although Piper them selves stopped making it some time ago.

This is not a duplicate of "Oldest aircraft still in production?".  This list could, in theory, include planes that were in production from 1920 - 2010, and thus are out of production but still would have a longer production run than any active plane.  Admittedly, practically, the answer ends up being the same for now.  But that may not always be true, and the questions are actually different so...don't close this one as a dupe please.

Probably the longest production run of any given craft is the Taylor Cub (more popularly known as the Piper Cub).  Which has, in one form or another, been built since about 1930, moving from manufacturer to manufacturer as the years go by.  There's been a recent resurgence in new manufacturers picking up the design lately because it now can be used in the LSA class of aircraft, created a few years back to encourage manufacturers to produce cheap, easy to fly, short range aircraft for recreational purposes.  CubCrafters is one such manufacturer, and here is their, very current, CarbonCub SS as an example of the type:

Source: https://upload.wikimedia.org/wikipedia/commons/2/25/CC11-160_Carbon_Cub.JPG, Author: FlugKerl2
The longest production run for an aircraft from a single manufacturer is the BeechCraft Bonanza (source).  Which has been in continuous production since 1947:

Source: https://i.sstatic.net/utPF8.jpg, Author: Bill Larkins
Another long production run is that of the Antonov An-2 which also started in 1947 and only ceased in 2001, though a variant, a turboprop version called the An-3, is still in production to this day:

Source: https://i.sstatic.net/YDvgG.jpg, Author: Sergey Ryabtsev
Other honerable mentions include the C-130 Hercules (in continuous production since 1954, even though only 2300 have been built).  And also, the Cessna 172 and Cessna 182.  Both started in 1956 and are still going strong today, with older airframes still being extremely common.
Also, it is probably worth pointing out the Uh-1 "Huey" Iroquois, which has been in production since 1959, and is probably the longest production run for a rotary wing aircraft:

Source: https://i.sstatic.net/jWVEg.jpg, Author: 111 Emergency

The curtains in that Bonanza are just... the icing on the cake! They have a very 50s feel to them, if only they were chintz


text:
How do airline pilots maintain night currency?
Does flying as an airline pilot under 14 CFR 121 provide different ways to maintain night currency? Suppose a pilot were to bid only day trips and end up flying only during the day for over 90 days. Under the recency of flight experience requirements in 14 CFR 61 that would mean a pilot might not be night current anymore but do 121 operator's op specs have some way around this?

Why does that seem reasonable? Sure the pilot knows what they need for currency, so they can bid accordingly and if they don't, the manager knows too and can reassign them.

@JanHudec I removed the words "seem reasonable."

The night currency requirements are in 14 CFR 61.57, which does provide exceptions for pilots flying for an air carrier:

Except as provided in paragraph (e) of this section, no person may act
  as pilot in command of an aircraft carrying passengers [at night]

61.57(e)(2) has this exception for part 121:

(2) This section does not apply to a pilot in command who is employed
  by a part 119 certificate holder authorized to conduct operations
  under part 121 when the pilot is engaged in a flight operation under
  part 91 or 121 for that certificate holder if the pilot in command
  complies with §§121.436 and 121.439 of this chapter.

121.436 has general PIC requirements for part 121, and 121.439 has recency requirements, which are 3 takeoffs and landings in the last 90 days in the same aircraft type, or an equivalent simulator.
So part 121 pilots need three takeoffs and landings in type in the last 90 days, with no specific requirement for them to be at night. That's all based on just reading the regulations, someone who's actually been there and done that could probably add some more information.


text:
Does an AQP line check conducted by a part 121 carrier count as a flight review and instrument proficiency check for part 91 operations?
If a pilot completes an AQP program with a part 121 airline, does completion of all training and passing the line evaluation also qualify as a flight review and instrument proficiency check for continued part 91 operations?

For what purposes? For qualifying for continued operation with that company under either §121 or §135? For operation outside the air carrier under §91? For §107 remote pilot certification? As asked this is rather broad—if not too broad. I'm not voting to close, but you should clarify.

It depends if the AQP qualifies as a pilot proficiency check.  Most likely the the outcome will be a pilot proficiency check.

§61.56   Flight review.

...
(d) A person who has, within the period specified in paragraph (c) of this section, passed any of the following need not accomplish the flight review required by this section:

(1) A pilot proficiency check or practical test conducted by an examiner, an approved pilot check airman, or a U.S. Armed Force, for a pilot certificate, rating, or operating privilege.


text:
Does being cleared for a visual approach supersede speed instructions from ATC? (FAA)
Does being issued a visual approach clearance supersede speed instructions from ATC under FAA rules?
Here's an example. Let's assume an aircraft is being vectored for landing at a large class B airspace airport like KATL and has been instructed to maintain 210 knots. About 5 miles from the final approach fix on the ILS the aircraft is cleared for the visual approach. Does the visual approach clearance supersede the instruction to maintain 210 knots?
This is assuming the visual approach clearance didn't include an additional speed instruction along with it.
Please include the regulation or FAA-published guidance.

I need to research to give the official links, but the answer is Yes, any approach clearance cancels any speed restrictions, unless restated in or after the approach clearance.

Sounds like there's a story behind this question. Not gonna ask you to tell it, though... :)

The AIM 4-4-12 says that an approach clearance cancels a speed restriction, but ATC can explicitly restate the restriction if needed:

Approach clearances supersede any prior speed adjustment assignments,
  and pilots are expected to make their own speed adjustments as
  necessary to complete the approach. However, under certain
  circumstances, it may be necessary for ATC to issue further speed
  adjustments after approach clearance is issued to maintain separation
  between successive arrivals. Under such circumstances, previously
  issued speed adjustments will be restated if that speed is to be
  maintained or additional speed adjustments are requested. Speed
  adjustments should not be assigned inside the final approach fix on
  final or a point 5 miles from the runway, whichever is closer to the
  runway

The ATC orders 5-7-1 give the controller guidance:

c. At the time approach clearance is issued, previously issued speed
  adjustments must be restated if required.
d. Approach clearances cancel any previously assigned speed
  adjustment. Pilots are expected to make their own speed adjustments to
  complete the approach unless the adjustments are restated


text:
A320 flap settings
Can anyone please explain about the 5 types of A320 flap settings - Up, Flap 1, Flap 2 and FULL. What do these mean? I've read that the A320 gives 2 outputs for the corresponding flap settings. What does that mean? And what is Flap 1+F?
A320:
Up - 0°/0°
Flap 1- 18°/0°
Flap 1+F- 18°/10°
Flap 2- 22°/15°
Flap 3- 22°/20°
Full- 27°/35

https://aviation.stackexchange.com/questions/42926/is-the-a320-family-allowed-to-take-off-with-the-flaps-retracted should help (within the first five hits on Google, by the way - and happy Christmas!)

@CptReynolds thx, that would help! Merry Christmas to you too!

1+F is a configuration with slats at the default position for the flap selector at "1" but with additional flaps extended (10°).
CONF 1+F is selected by positioning the flap lever from up to 1 when speed is below 100 kts or when the flap lever is moved from 2 to position 1.
In CONF 1+F the flaps automatically retract above 210 kts and the selected configuration changes to CONF 1.
As the earlier reply already mentioned the numbers are the slats/flap angles in degrees.
If you need more info I can highly recommend checking out the A320 FCOM e.g. available from http://www.smartcockpit.com/plane/AIRBUS/A320.html

The numbers you listed in the question are slats/flaps. So there is no difference in slat setting between config 1 and config 1+F, and no difference between config 2 and 3.
A good reference for this is "The A320 Podcast" on iTunes/Spotify. There's an episode dedicated to the flap and slat system which explains this really clearly, with the associated failures and procedures.

An A320 has 5 flap positions. 0, 1, 2, 3, and FULL.
Typically, 1+F or 2 and sometimes 3 is used for takeoff depending on operating factors like maneuvering speed, runway length, barometric pressure, weight, OAT, etc.
If you go for Flap 1+F on takeoff then you’d normally retract Flap to 0 at S speed (on PFD speed tape) but since the Airbus is a magical machine, if you fail to retract flap at S speed then auto retraction happens at 210 knots which means flaps are retracted and slats are still extended until you move that lever back to 0.
When in flaps 2, you’d normally choose flap 1 at F speed, then flap 0 at S speed. “O” is green dot speed as shown on the PFD speed tape as a green small dot which indicates the best lift to drag ratio speed or the engine out operating speed in clean config. Not to forget that GD speed is only displayed when in clean config (i.e. flap at 0).
Sorry if all that was overwhelming but good to know since you’re flying that machine on the simulator. Best of luck


text:
What aircraft equipment produced the three recordings of unidentified aerial phenomenon released by the US Pentagon in 2020? What does black signify?
I don't follow these things closely and my question is carefully scoped to ask ONLY about the type of aircraft equipment used and what the various gray levels in the images indicates.
In CNN "The Point" Video Barack Obama just said something very interesting about UFOs by Chris Cillizza, there are three snippets from videos which are described as follows:

In April 2020 when the Pentagon released three videos of what they called “unidentified aerial phenomenon” which amounted to a huge break in the government’s acknowledgement of their existence...

Below are cropped screenshots from my linked video, presumably from the three videos mentioned. The unidentified phenomena all appear as small black objects, sometimes with a lighter "halo" against gray backgrounds. If for example these are thermal infrared images, one might naively suppose that the black tone indicates a temperature far lower than anything else in the field of view.
However

Sometimes thermal infrared images are inverted so that cold is white and hot is black. Why are the clouds white and Australia black in weather satellite infrared images? and also this comment that images are sometimes inverted to make a bright object of interest easier to see when analyzing.
In some imaging systems (showing my age), tiny spots that are too bright and saturate a system can cause much larger and more prominent black areas to appear.

Question: What equipment produced the three recordings of unidentified areal phenomenon released by the US Pentagon in 2020? What does the black tone of the spot in question actually signify?

update: Screenshot from CNN's Retired US Navy Chief explains tech witnessed in UFO aircraft sighting around 00:13 shows a switch between opposite tone scales in the middle of a video clip, clearly the same clouds and object, just flipped in sign:

below left: Himawari-8 IR1 image from Central Weather Bureau showing hotter Australia represented with darker grayscale level (click for full size), right: Source "...illustrate(s) the artifact of a dark area around a bright area" (vidicon)

reminder that the question is carefully scoped to ask ONLY about the type of aircraft equipment used and what the various gray levels in the images indicates.

I think the photography questions might be as well or better suited to Photography.SE

@RonBeyer I'm guessing there are more pilots familiar with this particular kind of aircraft equipment than there are photographers. Only 6 out of my 54 questions have received zero answers so I have some faith in this community. If nothing shows up in a few weeks I'll reconsider. Thanks!

Note that even for a thermal image, that is in far infrared, which the FLIRs generally are, there is still significant amount of reflected sunlight complicating the matter.

I'm not sure there's an answer to this question in this particular video from this level-headed YouTube channel, but it does offer a lot of background on the FLIR and how its gimbaling motion and lens changes produce artifacts https://www.youtube.com/watch?v=U1di0XIa9RQ (also)

Putting this in a comment because frankly I find it confusing.  https://www.nytimes.com/2017/12/16/us/politics/unidentified-flying-object-navy.html shows a couple of different stills and videos.  You can see bottom left, below the Mach number, either BLK or WHT.  That should be Black Hot or White Hot respectively.  The top photo of the article would then show a "hot" UFO, but the bottom video would show a "cold" one.  At least as far as IR output; how that correlates with temperature can vary somewhat.

@ErinAnne I think you've got the answer! I can't read the NYTimes article (no more free articles this month) but searching found Navy Times March 12, 2018 Video shows apparent encounter between Navy pilot and UFO which leads to the answer in a YouTube video https://youtu.be/wxVRg7LLaQA?t=73 which confirms your statement. That's the answer to my question!

@uhoh ha well I'll try to remember to post it as an answer unless you want to go ahead and self-answer or something.  I'd look for an F-18 FLIR symbology reference too (I know Fravor flew F-18s, he was Nimitz CAG when they filmed the US version of "Carrier" in 2005) but would probably only find DCS ones

@ErinAnne I've bountied the question so please post so I can award it!

There are multiple UFO videos/photos being asked about, I think.  The clearest exhibition of the phenomena I know of are from the videos Dave Fravor released.  In this video/timestamp that @uhoh found, one of those "GO FAST" encounters is annotated by the "To The Stars Academy of Arts & Science" Youtube channel as from an ATFLIR targeting pod, and variously in BLK (Black Hot) and WHT (White Hot) mode.  I've taken the liberty of inexpertly combining three frames from the video in the following image:

The ATFLIR annotation at top
The BLK/WHT coloration annotation at left
The ATFLIR image itself is from a later timestamp, showing the target locked, and white, against a warmer surface (probably the sea; the ATFLIR is definitely reporting it's pointed 26 degrees down and to approx. 10 o'clock) while the ATFLIR is in BLK (Black Hot mode)

For this image, the UFO appears "cold."  Others, including the ones in the CNN screenshots in the question from a different encounter, are "hot"--see e.g. the NavyTimes article Video shows apparent encounter between Navy pilot and UFO that shows the uncropped image with a black target and BLK mode displayed.
A word of caution in interpreting IR imagery:  though the amount of infrared light coming from an object generally correlates well with its temperature, it isn't an absolute correlation.  See, for example, Figures 10 and 11 of Design and Preliminary Testing of the International Docking Adapter's Peripheral Docking Target, which I'll reproduce here:

Some materials on those docking targets, and the flight hardware that eventually went on to fly on the ISS, were specifically chosen to passively appear "cold" in infrared by being mostly-reflective in the measured bands.  Despite the docking targets being in approximate thermal equilibrium, the reflection of the sky over Houston is interpreted by the camera as starkly cold.
Not to say that you can't draw any conclusions on whether or not the objects in question are cooler or hotter than the background, just to say that it might not be entirely straightforward, especially from single still images.

This is excellent, thank you!

We know that at least one of the sensor platforms involved was a Raytheon ATFLIR, based on information from their own website:

Fast forward 13 years to Dec. 16, when the New York Times published a
report on the 2004 incident and a little-known Pentagon program that
tracked such reports from 2007 until at least 2012. Included was the
video of the possible UFO incident, which was captured by a Raytheon
Advanced Targeting Forward Look Infrared sensor, mounted under one of
the fighter jets.

The video was taken from an F/A-18 Super Hornet. This narrows down the recording equipment to either the ATFLIR or LITENING targeting pods. These pods perform essentially the same function. These pods can provide infrared imagery like that seen in the clip. Both pods are used on American military aircraft. I can't speak to the gray levels because I'm not familiar with the systems themselves, unfortunately. Hope this helps.

No d/v, but this is not really an answer, and should maybe be a comment.

@DigitalDracula it does indeed, thanks!


text:
Volume of air deflected down by a wing?
When trying to calculate, from first principles, the lift produced by the rotor of an autogyro, I have tried imagining the rotor as a circular flat wing set at an angle to the relative wind. The mass flow rate would be the volume times density of the swept 'air tube' deflected down per unit time. That mass flow rate, times the airspeed and the sine of the deflection, will give the lift... And if the gyro is flying s/l, you'll know that lift = weight, so that the deflection angle can be calculated.
But I don't know how to estimate the area of the base of that 'air tube'. In some papers, I've seen that it's taken as a circle with a diameter equal to the wingspan, but I find that somewhat arbitrary...
Is there a rational criterion for that area...?

If you use that circle with diameter = wingspan, the induced drag will be that of an elliptic wing. If you have less than optimal lift distribution, the circle will become a bit smaller.

I see... Well, the projection on the horizontal of a circular wing is an ellipse... Any reference for the derivation...? No problem if in German...

I did explain it here. The disk is perpendicular to the direction of flight, so there is no ellipse but really an idealized circular tube. Similar to the Froude hypothesis for propellers.

@PeterKämpf Thinking in your explanations, I have a doubt concerning the statement that for a wing with a less-than-optimal lift distribution, the section of the associated 'air tube' would be smaller. I think that, for those less-than-optimal wings, the section would be larger, as the vector sum of lift and drag will be larger for those sub-optimal wings, hence the total force on them will be higher, and the product mass flow rate x airspeed should be also higher too, something that –for all other things equal– would mean a larger tube section...

Less than optimum means that some sections deflect the air by more than others, and this unequal distribution is less efficient than a constant one. Using a smaller tube is only a mathematical trick to simulate wings with lightly loaded tips and I agree, this is not universally applicable. But making the tube bigger means that more air is deflected by less, creating lower drag. This only would be justified with a larger wing span. The simplification of a stream tube uses constant deflection over span which is not realistic for non-elliptical wings.

@PeterKämpf Thanks. I'm trying to adjust my 'paper results' to the real world of small gyro performance, and the area of the base of the 'stream tube' seemed to me a potentially modifiable parameter, that (in order that paper and real world might fit) should be corrected with a coefficient of approx 1,2...


text:
Does a plane have more lift while flying in a cloud? (All other things equal...)
Inside a cloud, the plane flies in a medium (air + water droplets) with a higher density than 'normal' air. If airspeed and AoA are the same, lift should be higher. It is really so...?

Are you certain that air parcels in clouds have a higher density than air parcels outside clouds? The opposite is often the case.

Also, and this might be critical to providing an accurate question, what do you mean by "all other things equal"? A cloud parcel of air cannot be equal to a non-cloud parcel of air. The factors that determine cloud formation (temperature, saturation, pressure, air mass stability, etc) necessarily render something not equal between the two parcels. What the difference is might determine the answer.

If we consider droplets "floating" in the air, they should hit the wings at the angle of attack and provide some lift. As they are more dense than the air, the lift should be more that just hitting the amount of air that could substitute the droplets.
There are also other effects like humid air is less dense and if the droplets are large enough to fall they should hit horizontal surfaces, pushing down. The overall effect depends on all these factors and may be complex to define.

No, higher humidity air is less dense than lower humidity air.  This is because water molecules are lighter than the nitrogen and oxygen they would replace in an airmass.
https://science.howstuffworks.com/nature/climate-weather/atmospheric/is-humid-air-heavier-than-dry-air.htm

any water vapor that gets added replaces either nitrogen or oxygen in our free-moving air. Nitrogen and oxygen make up the majority of our atmosphere, and they're displaced -- or evaporated -- when water takes their place in the air. And water vapor molecules are lighter than both nitrogen and oxygen. In other words, humid air is going to have less heavy nitrogen and oxygen -- and lighter hydrogen and oxygen -- in its place. Remember that they have the same number of molecules, but the air with water vapor is simply less dense

Humidity, therefore, is also an important factor in density altitude calculations.

This is a good thought, but by its very nature, at least some portion of the water present in a cloud is necessarily in a non-vapor form: either ice crystals or liquid water droplets. I'm not sure the water vapor density facts behind this answer apply to this question.

No, or at least not lift due to the airfoil (see below). The difference between the air inside of a cloud and that directly adjoining it will be relatively small. Also, the air inside a cloud is less dense in most cases than the surrounding air because it is warmer. The following diagram illustrates the effect:

The air outside the cloud is cooler and more dense. The air inside of a cloud, therefore rises, and outside of it, the air sinks. Any glider pilot experiences when passing beneath a cloud: the glider rises as it travels under the cloud, but as soon as it exits the area under the cloud, the glider sinks.
An aircraft will experience lift inside of a cloud, but this is due to rising thermal air currents, not lift granted by the wing's airfoil.

Yes, the air in the cloud is colder than the air BELOW it, but the air alongside of the cloud is colder than the air inside the cloud.

@jjack Thermal convection does not increase the angle of attack.

The cloud forms because the air is rising. The air is rising because it is less dense than surrounding air. As the air rises pressure decreases causing droplet formation. The droplet formation in turn causes the drop in temperature. Aerodynamic lift doesn't change. The AoA might go up briefly as you enter the cloud, but will quickly return to the trimmed state.  Flying at straight and level AoA through rising air causes positive climb rate. Same way the same airspeed with a tailwind yields a higher ground speed

@jjack When an aircraft is in a thermal, it rises along with the air. In other words, the aircraft is resting on the air, so if the air moves up as a body, it carries the aircraft with it. It works in reverse too. If the air is sinking, the aircraft will drop down.

@jjack You don't buy what?

Many clouds do not include an instability lifting force. Certain types of clouds do. Many cloud types include both updrafts and downdrafts. Other cloud types are stable with little or no associated up or down drafts. An aircraft my or may not experience some form of increase or decrease of lift inside a cloud.

@jjack Stability refers to an airmass's tendency to resist vertical displacement. Conversely, instability refers to an airmass's tendency toward vertical motion. Different types of clouds are related to stable or unstable airmasses with accompanying temperature and density dynamics.


text:
Can IAF and IF be in the same spot?
While looking over an ILS approach chart,
I found that there is a fix where says : (IF/IAF) ACZUP ...
And I was consfused whether the two IF and IAF can be in the same place, or the spot is used as IAF or IF in a certain occasion.
Could anyone tell me which is the right thing?

(IF = Intermediate Fix; IAF = Initial Approach Fix)

An IF is any fix between the IAF and FAF, so as a matter of logic, the same fix cannot be both an IAF and an IF at the same time. Therefore the label IAF/IF must mean it can be either one or, in some cases, both at different times.
If you are cleared to ACZUP for the straight-in approach, then ACZUP is the IAF only. You never pass it a second time, so it doesn't get a chance to be an IF.
If you are cleared to HABJE for the approach, then HABJE is the IAF and you do not pass over ACZUP, so it is neither IAF nor IF.
Otherwise, ACZUP is the IAF when you pass over it outbound, then you do the Hold in lieu of Procedure Turn (HiloPT), and then ACZUP is an IF when you pass over it again the second time inbound. The same would be true if it were charted as a PT instead of HiloPT.
In all three cases, MRVIN is the FAF due to the lightning bolt symbol.

Thank you for the comment! Is that an aviation knowledge or is it mentioned in the chart?

And I want to know if there is NO intermediate fix in the straight in approach...

@ StephenS Okay, I got it. I was just confused because of the fact 'No Intermediate fix' can happen.

@Student412 I've incorporated my previous comment into the answer; let me know if it's still unclear.


text:
How do power and thrust curves compare?
I would like to really understand why when talking about jets we care about thrust but when talking about propeller aircraft we care about power.
(THRUST)="Just a force"
(POWER)= work/time = force x velocity
I know that even though propeller aircraft are called "power producers" and not "thrust producers" as jets, any aircraft needs Thrust because that's what overcomes Drag. The difference is that a prop aircraft uses the Power of the engine to turn the prop while a turbine engine produces Thrust by itself.
I also know that total drag curve comes from induced drag (decreases with airspeed) and parasitic drag (increases with airspeed), so that is exactly what thrust needs to overcome. The total drag you get at a given airspeed is equivalent to the thrust you need to stay level. That makes a lot of sense.
I have no problem with jet theory when thinking about that and the Thrust required / Thrust available curves.
Why is it different for prop aircraft? Why is the power curve important? I mean, aerodynamically. Because I've seen that total drag curve is also the same as Power Required Curve and I don't understand why.
My first thought is:
Engine power is important because that is what the prop will use to rotate and generate Thrust. (Propeller efficiency is directly related with HP and other prop aerodynamic factors). However, even with the most powerful engine, if prop efficiency is poor because of the other factors, we won't get too much thrust and that's the only thing that matters. And one more time, thrust is what makes any aircraft fly forward, so total drag is thrust required, not power.
Vx speed is found where the greatest difference between thrust available and thrust required is. Greater excess thrust equals greater angle of climb. (Got it!) that's very logical for me.
But: Vy is where the greatest difference between Power Available and Power Required is found.
What?! Why?! How can the power the engine produces relate with airplane's aerodynamic?? It has nothing to do with that. Vy is best rate of climb, the airspeed at which the airplane will get faster to an altitude. I think that the reason should be related with a thrust/drag ratio.
On the other hand, I know that the Power Curve results from multiplying each thrust value times each airspeed value of the thrust curve. That's the fact, just let me explain why it doesn't make any sense to me...
Power=force x velocity, and we can apply that in different ways. If we were talking about electricity we could say Power is measured in Watts, and Watts= Voltage x Current. Because we are talking about prop aircraft, we should use a power unit like HP (for reciprocating engines). HP= Torque X RPM.
Multiplying the Thrust Curves to get Power Curves would mean that Power= Thrust x Airspeed
What kind of power are we getting when multiplying thrust times the airspeed at which the airplane is flying?
Wouldn't that be something like "Aircraft power = thrust x airpeed" or in other words "the amount of power needed to stop that airplane"? Nothing to do with the engine (which is really the reason why a prop airplane is called "power producer")
I'm very confused, even tired of not understanding the exact reasons behind this after reading what should be sufficient. I think I know what power and thrust are, but I just can't understand the way they relate, and to be more specific, how does excess power affect the aircraft flying in real life, aerodynamically. Why to talk about Power instead of Thrust in any case, jets or props. (It's clear there's a good reason not to, and that's what I want to finally understand). Maybe more realistic examples or comparisons instead of graphs and curves would help.
(But those are not a problem either in case you want to try.)
I'm a student pilot and I hope I was clear enough with my doubts and explanations.
(Also with English since Spanish is my native language)

Here would be a question which addresses how jet thrust changes with speed.

I'd say the difference between power output of an engine and thrust produced is the efficiency of the propeller or jet efflux plus losses due to friction, accessories etc

@quietflyer please can you avoid pseudo-answers in comments. If you want to post an answer please do so.

You've tangled many rather unrelated things together in your question. That makes it fairly difficult to answer comprehensively.

Re -- "Because I've seen that total drag curve is also the same as Power Required Curve and I don't understand why."  -- this isn't right; they aren't actually the same curves.  The question could be improved by fixing this.

Aircraft power curve
(a.k.a “polar”—a completely illogical name because somebody long ago plotted them in polar coordinates, which makes no sense, but the name stuck)
This is just the power balance of the aircraft: the power provided by the propulsion system minus the power taken away by drag.
You can divide the values by velocity (true airspeed) to get a force (thrust - drag) curve instead. But usually just the power curve is plotted and lines through origin are used to read the interesting force maxima.
Two useful observations can be made about meaning of the power and force here:

Excess power is linearly proportional to sustainable climb rate, simply by $P = m g v_v$ (where $P$ is power, $m$ is mass of the aircraft, $g$ is gravity and $v_v$ is vertical velocity).

Excess thrust is linearly proportional to sustainable angle of climb. Well, actually there is some tricky trigonometry involved, but at low angles (non-aerobatic aircraft only do very shallow climbs and descents, below about 10°) you can approximate $\sin x \approx x$ and $cos x \approx 1$ and say that $T = m g \gamma$ (where $T$ is thrust, $\gamma$ is flight path angle and $m$ and $g$ as above).

This is true independent of what kind of propulsion, if any, the aircraft has. Power curves work the same for propeller, jet and rocket aircraft and gliders alike. Just the function describing power available for given speed differs by the propulsion type.
Kinds of power
A propulsion system is a device that converts one form of energy to another with some efficiency. So in has input power ($P_{in}$)—which is the rate at which energy comes in—and output power ($P_{out}$)—which is the rate at which it provides the desired energy. They are related by efficiency ($\eta$: $P_{out} = \eta P_{in}$). The rest of the energy—because energy is always conserved—is wasted (often as heat, but in aircraft propulsion the energy given to the reaction mass, the air, is also wasted).
Powered aircraft usually use internal combustion engines consuming hydrocarbon fuel, so the input energy is fuel flow times heating value of the fuel.
The output power of the whole propulsion system is then the energy given to the aircraft (to replace the energy taken away by drag), and this is always thrust times velocity. They are related by the propulsive efficiency, which varies with velocity, and varies differently for different propulsion systems.
When propeller is used, you can split the system in two parts, with their own separate efficiencies: The engine burns fuel, drives the shaft, and wastes the rest of energy as heat of the exhaust gases. It's output power is the torque times shaft speed (the RPM). The propeller then takes this power as input and accelerates some air to produce thrust. But doing so it has to sacrifice some energy to that air (its kinetic energy increases), so its output power is lower than that of the engine.
Note to units

Power=force x velocity, and we can apply that in different ways. If we were talking about electricity we could say Power is measured in Watts, and Watts= Voltage x Current. Because we are talking about prop aircraft, we should use a power unit like HP (for reciprocating engines). HP= Torque X RPM.

Power is always the same dimension, and therefore its basic unit is always Watt. This is a derived unit composed as $\mathrm{W} = \frac{\mathrm{J}}{\mathrm{s}} = \frac{\mathrm{kg}\ \mathrm{m}^2}{\mathrm{s}^3}$. Power is voltage times current for electricity, and the units work out, because $\mathrm{V}$ is just $\frac{\mathrm{kg}\ \mathrm{m}^2}{\mathrm{A}\ \mathrm{s}^3}$ (and Ampere is a base unit).
For fuel, power is heat value, in $\frac{\mathrm{J}}{\mathrm{kg}}$, times flow rate, in $\frac{\mathrm{kg}}{\mathrm{s}}$ and we are at Watt again.
And of course torque in $\frac{\mathrm{kg}\ \mathrm{m}^2}{\mathrm{s}^2}$ times angular velocity in $\frac{1}{\mathrm{s}}$ gives you Watt again. (watch out here; it requires angular velocity, in radians per second, not revolutions per second, but radian works as dimensionless here (metre per metre), so this is prone to error by $2\pi$ that dimensional analysis won't catch).
A horsepower is not an independent unit, just a funny multiple of Watt. Multiple different multiples of Watt, actually, because there are some variants.
Engine power as function of velocity
Piston engine power grows with RPM, then plateaus (at the expense of declining efficiency) and then you hit the maximum permitted RPM.
A constant speed propeller allows running the engine at the optimal RPM, and it's efficiency only decreases slowly with airspeed. Therefore a piston engine with constant-speed propeller produces relatively constant power over the design range of the aircraft, while the thrust declines about inversely proportionally to velocity. Thus it makes most sense to give power as the main figure describing it.
With fixed pitch propeller, you are limited by not being able to reach the optimal RPM at low speeds, and by having to throttle the engine to avoid exceeding the maximum RPM at high speeds, so the power declines faster as you get away from some optimal design speed. That's why fixed propellers are only used on slower planes where it does not matter that much. The power curve is still flatter than the thrust curve (dividing by velocity), so it still makes more sense to give power as the main figure though.
With turbojet engines on the other hand the available thrust remains roughly the same. It first declines with speed somewhat similarly to a propeller, but then the ram pressure starts increasing the effective pressure ratio and the available thrust starts growing again. That's why jet engines are generally rated with thrust, not power. But it isn't really constant over speed, just like the power isn't really constant for a piston engine and propeller.
And this carries over to turbofan engines, that are actually somewhere between. Their thrust decreases with speed, though slower then for propellers, and their efficiency increases. It is an oversimplification to rate them with just static thrust, but it remains the primary value given.

could we rate "altitude energy use" rate as mg x delta h/s = mv$^2$/s for gliders?  Curious as to whether or not we eliminate "1/2" from the steady state energy/time formula. (It graphs as a square instead of a triangle).

@RobertDiGiovanni, which $\frac12$ factor do you mean? Potential energy is $E_p = m h g$, no $\frac12$ factors involved, and power needed for ascent (height grows up, so vertical speed is positive in climb) is simply $P_c = m g v_v$. There is no $t$ factor here, so the differentiation by it does not generate any non-unit factors. Only the kinetic energy gets $\frac12$ factor from going from the first derivative of position (velocity) to the second (acceleration), but the above discussion is for constant speed.

thanks for your response.  Your section on the interchangability of application for Power is very good.  I thought for gliding we could express use of altitude (delta h) per second as mv$^2$/s, which has the same units as fuel consumption J/s that you pointed out.

@RobertDiGiovanni, not just in gliding; always. Gliding is just a special case of engine power output being zero.


text:
What would FAA controllers be referring to when mentioning "code 0027?"
In the second video on "You Need To Hear These FAA Tapes From That Oregon UFO Incident That Sent F-15s Scrambling," you hear controllers mention a code 0027. If the target in question had a transponder, the content of the article is misleading. Does anyone know if code 0027 refers to something specific in ATC terms, or are they just referring to a transponder code?
Here's the full link to the article. http://www.thedrive.com/the-war-zone/18473/faa-recordings-deepen-mystery-surrounding-ufo-over-oregon-that-sent-f-15s-scrambling

Hearing the phrase in context would be very helpful but I counted 5 videos on that page, some of them over 30 minutes long. Which one are you referring to and when exactly in the video does the phrase appear?

@Pondlife it's in the second video, the one with the radar and ATC synced, right at the beginning.

@Pondlife Revised.

@757toga I removed that from the question, but still, the reality of the matter is mundane compared to the implication.


text:
Why is this type of rotor hub so rare...?
I wonder why is this type of helicopter rotor hub so rare... It is a very simple construction, having no flap or lead-lag hinges. A universal, constant-speed joint leaves the rotor free to find his position in space, and the only complication is the swashplate, essential in order to control cyclic and collective pitch. This type of hingeless rotor hub needs little maintenance and is relatively easy to build with standard materials... But it's nevertheless nowhere to be seen...

According to wiki, that's a "Doman LZ-1A (Sikorsky R-6 conversion)", for reference.

It's basically the teetering Bell/Robinson 2 blade rotor concept applied to a 4 blade rotor.  Like a teetering system, the gimballed rotor head accounts for most of the lead/lag/flapping, with whatever isn't accounted for absorbed by flexing of the long skinny root shafts.
With a 2 blade teetering system you have the weight of the machine hanging from a big cross bolt - pretty simple.  This system appears to be a bit complicated in the manner that the load of the machine is suspended from the rotor head, and that may have worked against it when other designers considered it.
The designer may have seen it as the advantages of a teetering system and an articulating multi blade system combined, whereas other designers saw mostly the disadvantages of the two.  An example: a teetering rotor, unlike an articulated one, can't tolerate hitting the teetering limits in flight because it puts huge bending loads on the mast (mast bumping) and makes the mast come apart.  This is why you never let a 2 blade machine go zero G.  I would expect this system could have a similar limitation, which a conventional articulated hub doesn't have.

Ha -- I found the same article about it.  That would be the only thing that I could think of as well.  That said, when you say that a 2-bladed teetering rotor is relatively simple (hanging from a bolt) and that this would be more complex...it would depend on the style of CV joint, since that's what the aircraft is hanging off of.  Some CV joints are as simple as a really well greased bolt.  My other thought for an answer to the OP would include a potential sensitivity to a localized hydraulic failure because of this.

Hard to say for sure but I think the weight of the machine is suspended by that wishbone shaped doohickey, a yoke with 4 arms creating a gimbal, that connects to the centre ring, not the CV joint, which I think transmits torque only.  That would make it a fairly complicated load bearing mechanism.  The control inputs are the same as any other helicopter, with a swashplate controlling blade rotation, so I would think that any hydraulics are to boost the mechanical control rods coming from the cyclic and a hydraulic failure would be the same as on any other machine with boosted controls.

Yeah, it is difficult to make it out...but I wouldn't think that the wishbone-thing (which seems to correspond to the portion of the tiltable ring that does not rotate) would react vertical loads.  It looks like it's part of some sort of radial bearing, not a thrust bearing that would react vertical loads, whereas the CV joint seems to connect to that dome structure that you can see in the picture of the LZ-1.  Going along with how the R22 rotor works, it'd make sense to me that torque and thrust would be transmitted at the same point.  Also, I misremembered how hydraulics were used here.

Well I'm convinced the CV joint is torque only because otherwise you wouldn't need any of that other gimbal/yoke stuff attached to the structure.  Remember that the teetering system just has the rotor mast which doubles as a drive shaft, and no other structural members.  A Robinson/Bell fuselage is effectively hanging by its transmission.  This thing has a whole structure that wouldn't be necessary if the CV joint was taking the weight of the machine.  CV joints on cars don't have to take tension loads.  I imagine one that can take several thousand pounds in tension would be pretty massive.

I was convinced that the CV joint was, effectively, the gimbal for the rotor head.  Where are the gimbal/yoke for the rotor head otherwise?  Could you label the drawings?  What I see are 4 blades on a teetering hub that tilts via the CV joint currently labeled in the drawing, which, similar to an R22, would transmit thrust forces from the rotor blades to the mast via a teetering hinge of some sort (sourced from the R22 maint. manual: https://robinsonheli.com/wp-content/uploads/2017/01/r22_mm_9.pdf).

I added an annotated version.  The red lines are what I think is the load path for machine's mass.  You see the Y shaped yoke with bearings connecting to a center ring.  Looks like a gimbal assembly to me. What's missing is how the yoke actually tilts.   I think a CV joint that can handle a tension load with the weight of the machine would be quite massive and heavy.

In this type of hub, the weight of the aircraft is taken by an spherical roller bearing. So it was in the first example, the Doblhoff gyro/helicopter. Description here: https://retromechanix.com/doblhoff-wnf-342-v4-evaluation-by-general-electric-1948/

This type of rotor is not common because flapping blades make the helicopter fly better. Flapping vertically allows the blades to reduce angle of attack as they advance/rise and increase angle of attack as they retreat/fall, evening out lift in concert with cyclic feathering. Flapping horizontally damps vibration and eases acceleration/deceleration.

Flapping and lead-lag hinges just allow the tip-path plane to orient itself in space in response to aerodynamic forces, working as a big universal joint. The same effect, with simpler components, may be achieved with the 'floating hub' used by the Doman choppers (and also by the pioneering Doblhoff gyro-helicopter).

there is in fact a german-made helo that uses a "rigid rotor" like this, but I can't remember its name...

@niels nielsen  You probably mean the Bo105. It's true that it uses a hingeless rotorhead, but it isn't a 'floating hub' rotor. There are several hingeless designs around, based upon elastic materials, but it isn't the same thing...

that's the one. -NN


text:
To which security domain does the Cabin Intercommunication Data System (CIDS) belong?
I've read a lot about Cabin Intercommunication Data System (CIDS) functionalities and features, but yet it is not really clear to me if CIDS is part of the Airline Information Services Domain (AISD), or the Passenger Information & Entertainment Services Domain (PIESD).
Does CIDS receive information/messages from the Aircraft Communications and Reporting System (ACARS), and if so, through which ARINC protocol, and does it have any kind of connection to the Aircraft Control Domain (ACD)?

I'll readily admit that I'm not familiar with these systems, but might the answer to this not depend on the specific aircraft (at least model) in question? It might help focus answers if you specify an aircraft model that you are interested in.

@MichaelKjörling Well I'm not sure it could help since the domains are from the ARINC 664 Part 5 that is a standard and ACARS and CIDS are somehow a standard presence on aircrafts.

As the commentators above pointed out, there are some issues with your question due to the concepts involved.
As @Bluephoenix mentioned, the security domains are defined in ARINC 664 Part 5. These domains specifically apply to ARINC 664 networks, though the concepts can be applied to other networks.
You also reference a Cabin Intercommunication Data System (CIDS) which is an Airbus specific system in the A320 series -- which does not have an A664 network. But we can look at its functions to see which domain would apply if they were implemented on an A664 network.
The A664P5 descriptions are:

Part 5 describes the objectives, characteristics, services and
  functions that should be satisfied and provided by the aircraft
  computing network as a whole. Their allocation to specific networks
  and devices is the responsibility of aircraft network designers and
  specified by other documents.
The Aircraft Control Domain (ACD) and Airline Information Services
  Domains (AISD) can be divided into sub-domains.
The ACD can be divided into two Sub-domains:
  • Flight and Embedded Control System Sub-domain, where the aircraft is
  controlled from the flight-deck
  • Cabin Core Sub-domain, which provides environmental functions
  dedicated to cabin operations, such as environmental control,
  passenger address, smoke detection, etc.
The AISD can be subdivided into two sub-domains:
  • Administrative Sub-domain, which provides operational and airline
  administrative information to both the flight deck and cabin
  • Passenger Support Sub-domain, which provides information to support
  the passengers
Subsystems within the Passenger Information and Entertainment Services
  Domain (PIESD) could be supplied by different suppliers.
Passenger Owned Devices (POD) can be supported by onboard networks.
  The Passenger Owned Devices Domain (PODD) is constituted of only those
  devices that passengers may bring on board the aircraft.

Reviewing the CIDS documentation of functions and comparing it to the A664P5 definitions, it would appear that the CIDS has functions within the ACD Cabin Core Sub-domain (environmental control, passenger address, smoke detection, etc.) and AISD Administrative Sub-domain (operational and airline administrative information to both the flight deck and cabin.)
To your last question, ACARS is a legacy communications link designed to support Airline Operational Control (AOC) and Airline Administrative Control (AAC) messaging.  With advances in data link systems and applications, ACARS can refer to either the AOC messaging (most correctly) or the entire data link system (not so correct) which also includes Air Traffic Control (ATC) messaging.
The ATC portion of data link falls under the ACD. AOC and AAC messaging fall into the AISD. In the case of the A320, the CIDS does not have a direct interface to the AOC/AAC messaging. These messages (manifests, gate assignments, etc.) are printed out in the cockpit and passed to the cabin crew as needed. On some newer aircraft, this may be accessible by the cabin crew via a terminal.

Thanks for your good answer, but I still have a doubt about your placing of the CIDS both in ACD and AISD. The ACD should be a closed domain and so how can a component, like CIDS, have some parts of it in a closed domain and the remainings in a not-completely-closed domain?

I didn't put it there, the AEEC (the writers of A664) put it there. And if you note, ACD has two sub-domains - one for flight deck and one for cabin. The functions included in the Cabin Core sub-domain are aircraft control functions - environmental control, fire detection, PA, doors - and can be monitored and controlled from the flight deck and the FA station. The domains provide a firewall that allows the PA to override the IFE but blocks the IFE from accessing any control systems.